Math and Arithmetic

Word Brain Teasers

# 24 b b b in a p?

###### Wiki User

###### August 17, 2008 7:33PM

24 black birds baked in a pie. From the nursery rhyme Sing A Song Of Sixpence.

## Related Questions

###### Asked in Math and Arithmetic, Probability

### How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?

There are symbols missing from your question which I cam
struggling to guess and re-insert.
p(a) = 2/3
p(b ??? a) = 1/2
p(a ∪ b) = 4/5
p(b) = ?
Why use the set notation of Union on the third given probability
whereas the second probability has something missing but the "sets"
are in the other order, and the order wouldn't matter in sets.
There are two possibilities:
1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2
→ p(a) + p(b) = p(a ∪ b) + p(a ∩ b)
→ p(b) = p(a ∪ b) + p(a ∩ b) - p(a)
= 4/5 + 1/2 - 2/3
= 24/30 + 15/30 - 20/30
= 19/30
2) The second and third probabilities are probabilities of
"given that", ie:
p(b|a) = 1/2
p(a|b) = 4/5
→ Use Bayes theorem:
p(b)p(a|b) = p(a)p(b|a)
→ p(b) = (p(a)p(b|a))/p(a|b)
= (2/3 × 1/2) / (4/5)
= 2/3 × 1/2 × 5/4 = 5/12

###### Asked in Probability

### Give the example of why probabilities of A given B and B given A are not same?

Let's try this example (best conceived of as a squared 2x2 table
with sums to the side). The comma here is an AND logical operator.
P(A, B) = 0.1 P(A, non-B) = 0.4 P(non-A, B) = 0.3 P(non-A, non-B) =
0.2 then P(A) and P(B) are obtained by summing on the different
sides of the table: P(A) = P(A, B) + P(A, non-B) = 0.1 + 0.4 = 0.5
P(B) = P(A,B) + P(non-A, B) = 0.1 + 0.3 = 0.4 so P(A given B) = P
(A, B) / P (B) = 0.1 / 0.4 = 0.25 also written P(A|B) P(B given A)
= P (A,B) / P (A) = 0.1 / 0.5 = 0.2 The difference comes from the
different negated events added to form the whole P(A) and P(B). If
P(A, non-B) = P (B, non-A) then P(A) = P(B) and also P(A|B) =
P(B|A).

###### Asked in Probability

### Definition of additive law in probability?

This has to do with the union of events. If events A and B are
in the set S, then the union of A and B is the set of outcomes in A
or B. This means that either event A or event B, or both, can
occur.
P(A or B) = P(A) + P(B) - P(A and B)
**P(A and B) is subtracted, since by taking P(A) + P(B), their
intersection, P(A and B), has already been included. In other
words, if you did not subtract it, you would be including their
intersection twice. Draw a Venn Diagram to visualize.
If A and B can only happen separately, i.e., they are
independent events and thus P(A and B) = 0, then,
P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - 0 = P(A) +
P(B)

###### Asked in Math and Arithmetic

### What is the Formula for odds for either of two events?

If A and B are mutually exclusive event then Probability of A or
B is P(A)+P(B).
If they are not mutually exclusive then it is that minus the
probability of the P(A)+P(B)
That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it
is clear that if they are mutually exclusive, P(A and B)=0 and we
have the first formula.

###### Asked in Probability

### What is multiplication rule in probability?

Given two events, A and B, the conditional probability rule
states that
P(A and B) = P(A given that B has occurred)*P(B)
If A and B are independent, then the occurrence (or not) of B
makes no difference to the probability of A happening. So that
P(A given that B has occurred) = P(A)
and therefore, you get
P(A and B) = P(A)*P(B)

###### Asked in Statistics, Colleges and Universities, Probability

### What is addition theorem of probability?

Consider events A and B. P(A or B)= P(A) + P(B) - P(A and B)
The rule refers to the probability that A can happen, or B can
happen, or both can happen together. That is what is stated in the
addition rule.
Often P(A and B ) is zero, if they are mutually exclusive. In
this case the rule just becomes P(A or B)= P(A) + P(B).

###### Asked in Alex Rider

### Where can you buy Alex rider the Gemini project?

Well for one you cant buy Alex Rider the Gemini Project because
it doesn't exist what your after is Alex Rider Point Blanc and as
for that you can buy it in many places like amazon or got to the
libary and get it
:) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
:) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
:D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D
:D :D :D :D :D :D :D
:p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p
:p :p :p :p :p :p :p :p :p
:o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o
:o :o :o :o :o :o :o :o :o
:b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b
:b :b :b :b :b :b :b :b :b
:x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x
:x :x :x :x :x :x :x :x :x :x :x :x
:) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
:) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

###### Asked in History of the United States, American Revolution, Declaration of Independence

### Prove if A and B are independent events than A' and B' are independent events?

first prove *: if A intersect B is independent, then A intersect
B' is independent. (this is on wiki answers)
P(A' intersect B') = P(B')P(A'|B') by definition
= P(B')[1-P(A|B')] since 1 = P(A) + P(A')
= P(B')[1 - P(A)] from the first proof *
= P(B')P(A') since 1 = P(A) + P(A')
conclude with P(A' intersect B') = P(B')P(A') and is therefore
independent by definition.
***note*** i am a student in my first semester of probability so
this may be incorrect, but i used the first proof* so i figured i
would proof this one to kinda "give back".

###### Asked in Statistics, Probability

### Why is the probability of intersection of two events subtracted in the sum of the events?

The short answer is that if you don't do so, the intersection is
double counted. The longer answer follows:
Suppose you have two events A and B which overlap. Then,
Event A = (only A happens) and (both A and B happen).
These two are mutually exclusive so that P(A) = P(A only) +
P(both).
Similarly, P(B) = P(B only) + P(both).
Now, the event that A or B happen is the event that only A
happens or only B happens or both happen.
That is, P(A or B) = P(A only) + P(both) + P(B only)
Adding and subtracting P(both) gives
P(A or B) = P(A only) + P(both) + P(B only) + P(both) -
P(both)
Now, the first two are P(A) and the next two are P(B)
So P(A or B) = P(A) + P(B) - P(both).